Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, m, q, par[N], high[N], up[N][20];
vector<int> ar[N], adj[N], dp[N][20];
void make_par(int v, int pr, int h) {
  par[v] = pr;
  high[v] = h;
  for (int i : adj[v])
    if (i != pr) make_par(i, v, h + 1);
}
vector<int> merge(vector<int> v1, vector<int> v2) {
  vector<int> res;
  res = v1;
  for (int i : v2) res.push_back(i);
  sort(res.begin(), res.end());
  res.resize(distance(res.begin(), unique(res.begin(), res.end())));
  while (res.size() > 10) res.pop_back();
  return res;
}
vector<int> solve(int &v, int dis) {
  if (!dis) return ar[v];
  for (int i = 20; i >= 0; i--)
    if (dis & (1LL << i))
      return merge(dp[v][i], solve(up[v][i], dis ^ (1LL << i)));
}
int go_up(int &v, int dis) {
  if (!dis) return v;
  for (int i = 20; i >= 0; i--)
    if (dis & (1LL << i)) return go_up(up[v][i], dis ^ (1LL << i));
}
int LCA(int u, int v) {
  if (high[u] > high[v])
    u = go_up(u, high[u] - high[v]);
  else if (high[u] < high[v])
    v = go_up(v, high[v] - high[u]);
  if (u == v) return u;
  int st = 0, ed = n;
  while (ed - st > 1) {
    int mid = (st + ed) / 2;
    if (go_up(u, mid) == go_up(v, mid))
      ed = mid;
    else
      st = mid;
  }
  return go_up(u, ed);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = 0; i < m; i++) {
    int a;
    cin >> a;
    a--;
    ar[a].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    sort(ar[i].begin(), ar[i].end());
    while (ar[i].size() > 10) ar[i].pop_back();
  }
  make_par(0, 0, 0);
  for (int i = 0; i < n; i++) {
    vector<int> vec, vec2;
    for (int j : ar[i]) vec.push_back(j);
    for (int j : ar[par[i]]) vec2.push_back(j);
    dp[i][0] = merge(vec, vec2);
    up[i][0] = par[i];
  }
  for (int j = 1; j < 18; j++)
    for (int i = 0; i < n; i++) up[i][j] = up[up[i][j - 1]][j - 1];
  for (int j = 1; j < 18; j++)
    for (int i = 0; i < n; i++)
      dp[i][j] = merge(dp[i][j - 1], dp[up[i][j - 1]][j - 1]);
  for (int i = 0; i < q; i++) {
    int u, v, c, lca;
    cin >> u >> v >> c;
    u--;
    v--;
    lca = LCA(u, v);
    vector<int> vec =
        merge(solve(u, high[u] - high[lca]), solve(v, high[v] - high[lca]));
    cout << min(c, (int)vec.size()) << " ";
    for (int j = 0; j < min(c, (int)vec.size()); j++) cout << vec[j] + 1 << " ";
    cout << endl;
  }
}